Transcript Mechanics of Materials II Lecture # 06(B)

Mechanics of Materials II
UET, Taxila
Lecture No. (6)
Final form of hoop stress
Longitudinal stress final form


This means that:
Longitudinal stress = ½ of the hoop
stress

Or:

Hoop stress = 2 Longitudinal stress
From Hooke’s Law
From Hooke’s Law:
Circumferential strain H ) or θ )=
Change in Diameter
(c) Change in internal volume
Volumetric strain = change
in volume /original volume
Change in volume =
volumetric strain x original
volume
Volumetric strain =
One longitudinal strain + 2 Diametral
strains
Final form for change of volume equation:
Change in volume = volumetric strain x
original volume
if “V” is the original volume
Thin rotating ring or
cylinder
Thin rotating ring or cylinder

Consider a thin ring or cylinder as
shown in Next Figure
pc
Rotating thin ring or thin cylinder.
subjected to a radial
pressure pc caused by
the centrifugal effect of its
own mass when rotating.
 The centrifugal effect on a
unit length of the
circumference is:

Pc = m
2
 r

Where:

pc is the pressure,
m is the mass,

 is the angular velocity (rad/s)



and
r is the inner radius of the cylinder.

Thus, considering the equilibrium of
half the ring shown in the figure:
c
c


where F is the hoop tension set up
owing to rotation.
The cylinder wall is assumed to be so
thin that the centrifugal effect can be
taken to be constant across the wall
thickness then:
c
Hoop stress owing to rotation

This tension is transmitted through the
complete circumference and therefore
is restricted by the complete crosssectional area then:
where A is the cross-sectional area of
the ring.
Now with unit length assumed,
m/A is the mass of the ring
material per unit volume, i.e.
the Density  .
 Then hoop stress owing to
rotation:

Thin spherical shell
under internal
pressure
Thin spherical shell under internal pressure

Because of the symmetry
of the sphere the stresses
set up owing to internal
pressure will be two
mutually perpendicular
hoop or circumferential
stresses of equal value
and a radial stress.

As with thin cylinders
having thickness to
diameter ratios less than
1 : 20, the radial stress is
assumed negligible in
comparison with the
values of hoop stress set
up.

The stress system is
therefore one of
equal biaxial hoop
stresses.

Consider, therefore, the equilibrium of
the half-sphere shown in Next Figure:
Half of a thin sphere subjected to internal
pressure showing uniform hoop stresses acting
on a surface element.
For Sphere
For Sphere
Change in internal
volume for sphere
Change in internal volume
As for the cylinder:
change in volume = original
volume x volumetric strain

But:
 volumetric strain = sum
of three mutually
perpendicular strains (in
this case all equal)
Sphere volumetric strain
By considering V as the original volume
Vessels subjected
to fluid pressure
Vessels subjected to fluid pressure

If a fluid is used as the
pressurisation medium
the fluid itself will change
in volume as pressure is
increased.

This must be taken into
account when calculating the
amount of fluid which must be
pumped into the cylinder in
order to raise the pressure by a
specified amount, the cylinder
being initially full of fluid at
atmospheric pressure.
Bulk Modulus

The bulk modulus (K)
of a substance
measures the
substance's resistance
to uniform
compression.
Definition of bulk modulus

It is defined as the
pressure increase
needed to effect a
given relative decrease
in volume.

Now the bulk modulus of a fluid
is as follows:
bulk modulus “K”
=volumetric stress/ volumetric
strain
where, in this case, volumetric
stress = pressure p
 And Volumetric strain =


Change in volume / original
volume
= V/V

Then:

Change of volume under pressure =
pV / k

The extra fluid required to
raise the pressure must,
therefore, take up this
volume together with the
increase in internal
volume of the cylinder
itself.

Then: extra fluid required to raise
cylinder pressure by p =
For Sphere

Similarly, for spheres, the extra fluid
required is